Can you find all the zeros of p(x)

x^3 - 14x^2 + 53x - 40=0
(x-1)(x-5)(x-8)=0
x = 1, 5 and 8

If x = 1 is a zero of P(x), then (x - 1) is a factor. You're supposed to do polynomial division to divide P(x) by (x - 1). Then you'll have a quadratic, which you can solve with factoring or the quadratic formula.

Obviously you've just had a unit on polynomial division, so apply the techniques you learned there.

x^3-14x^2+53x-40=(x-1)(x-5)(x-8)
x1=1
x2=5
x3=8